Problem: Let $n\ge 3$. Let $x_i \ge 0, \forall i$. Prove that
$$(n-1)\sqrt{1+\dfrac{1}{n-1}\left(n\sum_{i=1}^{n}x^2_{i}-\left(\sum_{i=1}^{n}x_{i}\right)^2\right)}
+\prod_{i=1}^{n}x_{i}\ge\sum_{i=1}^{n}x_{i}.$$
Proof:
By Vasc's Equal Variable Theorem [1, Corollary 1.8], we only need to prove the case when
either $x_1 = 0$ or $0 < x_1 \le x_2 = x_3 = \cdots = x_n$.
- $x_1 = 0$:
It suffices to prove that
$$(n-1)\sqrt{1+\dfrac{1}{n-1}\left(n\sum_{i=2}^{n}x^2_{i}-\left(\sum_{i=2}^{n}x_{i}\right)^2\right)}
\ge\sum_{i=2}^{n}x_{i}$$
that is
$$\frac{(n - 1)^2}{n} + (n - 1)\sum_{i=2}^{n}x^2_{i} \ge \left(\sum_{i=2}^{n}x_{i}\right)^2$$
which is true since $(n - 1)\sum_{i=2}^{n}x^2_{i} \ge \left(\sum_{i=2}^{n}x_{i}\right)^2$ by AM-QM.
- $0 < x_1 \le x_2 = x_3 = \cdots = x_n$:
Let $x_1 = a, x_2 = b$. It suffices to prove that, for all $0 < a \le b$,
$$(n - 1)\sqrt{1 + (b - a)^2} + a b^{n - 1} \ge a + (n - 1) b$$
that is
$$\frac{\sqrt{1 + (b - a)^2} - b}{a} \ge \frac{1 - b^{n - 1}}{n - 1}.$$
It is easy to prove that $x \mapsto \frac{1 - b^x}{x}$ is non-increasing on $[1, \infty)$.
Thus, it suffices to prove the case when $n = 2$, that is
$$\frac{\sqrt{1 + (b - a)^2} - b}{a} \ge 1 - b$$
which is true (easy to prove).
We are done.
Reference
[1] Vasile Cirtoaje, “The Equal Variable Method”, J. Inequal. Pure and Appl. Math., 8(1), 2007.
https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06.pdf
